How can we express the homotopy pullback of the tensor product diagram (more importantly, its homology) in terms of the homotopy pullbacks of the first two diagrams?

Now I'll give some motivation, because maybe someone knows the answer to the question the above seeks to find. I am interested in whether there is a Kunneth formula for (mainly ordinary) differential cohomology. The model of ordinary differential cohomology I am currently working with is as the homotopy pullback of

where the subscript $\mathbb{Z}$ denotes forms with integral periods. Taking homology commutes with taking homotopy pullbacks since we can replace anything by something quasi-isomorphic without affecting the quasi-isomorphism type (so I assume...), so Somehow we can from the above description produce a chain complex whose homology is ordinary differential cohomology. That is, the above diagram comes from

which has as homotopy pullback the chain complex $\Omega^k_\mathbb{Z}\times C^{k-1}_\mathrm{dR}\times C^k$ with differential

$(F,\mu,c)\rightarrow (0,F-c+d\mu,dc)$.

We have Eilenberg-Zilber q-isoms $\Omega_\mathbb{Z}(M\times N)\rightarrow\Omega_\mathbb{Z}(M)\otimes \Omega_\mathbb{Z}(N)$ etc for each of the three chain complexes above such that the diagram for $M\times N$ becomes the tensor product of the diagrams for $M$ and $N$. I expect that this implies something nice about the pullback, but I am not sure what.

I would also appreciate any answers using another model of differential cohomology (eg. Deligne cohomology), being ultimately interested in geometrical application, but I would like to understand this piece of homological algebra.

Taking homology does not commute with taking homotopy pullbacks. A complex over a ring of dimension $\leq 1$, e.g. a field or $\mathbb{Z}$, is quasi-iso to its cohomology, but a map of complexes is not in general quasi-iso to the induced map in cohomology. It may be in your case since the target consists of $\mathbb{R}$-vector spaces, but there's still a problem in this case: the quasi-isos cannot be made natural in $M$.
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Fernando MuroSep 24 '12 at 19:40

@Fernando I see. Thank you. I am certain that the chain complex I describe has the homology I want, so it must be something nice in this case about how the three complexes fit together.
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Ryan ThorngrenSep 24 '12 at 19:57